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(B) Given the vector equation: $(\frac{7}{3}+\beta) \hat{i}-\hat{j}+(\alpha+\gamma) \hat{k}=\frac{5}{3}(\alpha \hat{i}+\hat{j}-\hat{k})+\beta(2 \hat{j

Vedclass · Accepted Answer

$12$

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(B) Given the vector equation: $(\frac{7}{3}+\beta) \hat{i}-\hat{j}+(\alpha+\gamma) \hat{k}=\frac{5}{3}(\alpha \hat{i}+\hat{j}-\hat{k})+\beta(2 \hat{j}+\hat{k})+(\hat{i}+\gamma \hat{j}+3 \hat{k})$.Expanding the right side:$(\frac{7}{3}+\beta) \hat{i}-\hat{j}+(\alpha+\gamma) \hat{k}=(\frac{5}{3} \alpha+1) \hat{i}+(\frac{5}{3}+2 \beta+\gamma) \hat{j}+(-\frac{5}{3}+\beta+3) \hat{k}$.Comparing the coefficients of $\hat{i}, \hat{j}, \hat{k}$ on both sides:$1) \frac{7}{3}+\beta = \frac{5}{3} \alpha+1 \Rightarrow 5 \alpha-3 \beta=4$.$2) -1 = \frac{5}{3}+2 \beta+\gamma \Rightarrow 2 \beta+\gamma=-\frac{8}{3}$.$3) \alpha+\gamma = -\frac{5}{3}+\beta+3 \Rightarrow \alpha-\beta+\gamma=\frac{4}{3}$.From $(2)$, $\gamma = -\frac{8}{3}-2 \beta$. Substituting into $(3)$:$\alpha-\beta+(-\frac{8}{3}-2 \beta) = \frac{4}{3} \Rightarrow \alpha-3 \beta = 4$.This is the same as $(1)$. We have two independent equations for three variables. However, solving the system gives $\alpha=0, \beta=-\frac{4}{3}, \gamma=0$.Substituting these values into the expression:$5 \alpha-9 \beta+13 \gamma = 5(0)-9(-\frac{4}{3})+13(0) = 3(4) = 12$.

If $\alpha, \beta, \gamma$ are real numbers such that $(\frac{7}{3}+\beta) \hat{i}-\hat{j}+(\alpha+\gamma) \hat{k}=\frac{5}{3}(\alpha \hat{i}+\hat{j}-\hat{k})+\beta(2 \hat{j}+\hat{k})+(\hat{i}+\gamma \hat{j}+3 \hat{k})$, then $5 \alpha-9 \beta+13 \gamma=$

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